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This one is giving me some truble to modify correctly to allow me get $$\lim\limits_{x\to+\infty} \frac{(x + (x +(x)^{\frac 12})^{\frac 12})^{\frac 12}}{(x+1)^{\frac 12}} = 1$$

I keep on trying fatoring and turning it around but i can't seem to get away from $0/0$

(i gotta admit, i'm pretty bad at limits, i would welcome any recommendations for some good site for learning limits and possibly some solved problems to get experience on)

zwim
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Zerg Overmind
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3 Answers3

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HINT:

$$\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt{x+1}}=\dfrac{\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}}{\sqrt{1+\dfrac1x}}$$

lab bhattacharjee
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  • How did you factor out the x out of there? Wouldn't you need x^2 to factor it out? This ending up with 1/x ? I mean for example the bottomis x+1 thus to factor the x out of the square root you would need to have x^2 right? That means x ( 1/x + 1/x^2)^1/2 no? – Zerg Overmind Mar 16 '17 at 17:12
  • @ZergOvermind, Factored out $\sqrt x=\sqrt{\sqrt{x^2}}=\sqrt{\sqrt{\sqrt{x^4}}}$ – lab bhattacharjee Mar 16 '17 at 17:14
  • I indeed am foolish. I kept on thining about impossibility of getting it out as addition, i didn't even think of simply factoring it out as a root. Thank you for help. – Zerg Overmind Mar 16 '17 at 17:36
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Squeezing seems the most natural and straightforward approach to me: $$\sqrt{x}\leq \sqrt{x+\sqrt{x+\sqrt{x}}} \leq \sqrt{x+\sqrt{2x}}\leq\sqrt{x}\left(1+\frac{1}{\sqrt{2x}}\right)$$ and both $\lim_{x\to +\infty}\frac{\sqrt{x}}{\sqrt{x+1}}$ and $\lim_{x\to +\infty}\frac{\sqrt{x}}{\sqrt{x+1}}\left(1+\frac{1}{\sqrt{2x}}\right)$ clearly equal $1$.

Jack D'Aurizio
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The method is always the same, spot the dominant factor, divide everthing else by it (so this everything else goes to $0$) and either the limit appears naturally either you can now perform taylor expansions with things that are $\ll 1$.

This is what lab bhattacharjee did in his answer.


But notice also that you can spot the result without going too formal with the $o(\cdot)$ and expliciting the infinitesimal quantities, by using equivalents when there is no doubt about the dominance relation.

In our case notice that $\displaystyle{\frac{\sqrt x}{x}=\frac{1}{\sqrt x}\to 0}$ when $x\to+\infty$.

This means that $\sqrt x=o(x)$ but more importantly that $(x+\sqrt x)\sim x$


So in the expression you gave, we can have a chain reaction

$\sqrt{x+\sqrt{\vphantom{|} x+\sqrt{\vphantom{|}x}}}\sim\sqrt{\vphantom{|}x+\sqrt{\vphantom{|} x}}\sim\sqrt{x}$

For the denominator $\sqrt{\vphantom{|}x+1}\sim\sqrt x$

And finally you get $f(x)\sim\frac{\sqrt x}{\sqrt x}\sim1$ which is the searched limit.

zwim
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